A heat engine has an efficiency $ \eta $. Temperatures of source and sink are each decreased by $ 100 \mathrm{~K} $. Then, the efficiency of the engine.

Question

A heat engine has an efficiency $ \eta $. Temperatures of source and sink are each decreased by $ 100 \mathrm{~K} $. Then, the efficiency of the engine., Increases, Decreases, Remains constant, Becomes $ 1 $

MARKS App · Accepted Answer

η = 1 - T 2 T 1 = T 1 - T 2 T 1 where T 1 and T 2 are the temperatures of a source and sink respectively. When T 1 and T 2 both are decreased by 100 K each, (T 1 - T 2 ) stays constant. T 1 decreases. ∴ η increases.

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